Symmetric Group and Alternating Subgroup Prove that $A_{4}$ is the only subgroup of $S_{4}$ of order $12$.
-So far I know that the best way to prove this is by contradiction, assuming that there exists another subgroup in $S_{4}$ of order $12$:
Suppose $H$ is an subgroup of $S_{4}$ of order $12$. Since the order of $S_{4}$ is $24$, the index of $H$ is $24/12=2$.
What next?
 A: Since a subgroup of $H$ (of order 12) in $S_4$ is of index 2, it must be normal. 
This means it must be a union of conjugacy classes: if an element belongs to $H$ so does all of its conjugates. 
In $S_4$ all of the transpositions are conjugate: $\{ (12),(13),(14),(23),(24),(34)\}$ (6 of them). The 3 cycles are all conjugate: $\{ (123),(132),(124),(142),(134),(143),(234),(243)\}$ (8 of them). The 4 cycles are all conjugate: $\{ (1234),...\}$ ($(4-1)!=6$ of them). The disjoint pairs of transpositions are conjugate: $\{(12)(34),(13)(24),(14)(23)\}$ (3 of them). Finally, the identity is just conjugate to itself. 
We get: 6+8+6+3+1=24.
Now if some element is in a normal subgroup, everyone of its conjugates must be in there too. So a normal subgroup can be decomposed into the disjoint union of conjugacy classes. Keeping in mind, the identity must belong to any subgroup, how can we get 12 out of these numbers: 1,3,6,6,8? Just one way: 1+3+8. This exactly corresponds to the even permutations. :)
Another Approach Show that any subgroup of $S_n$ is either all-even or half-even/half-odd (hint: if not all-even, pick an odd member and use this to get a bijection between the even and odd elements in the subgroup). 
Next, assume that we have a subgroup of order $n!/2$ which isn't $A_n$, call this $H$. Then $H$ is half-even/half-odd. Therefore, $A_n \cap H$ is a subgroup of $A_n$ of index 2. 
But $A_n$ doesn't have subgroups of index 2. $A_2$ and $A_3$ are of orders 1 and 3 so they can't have a subgroups of orders 0.5 and 1.5 (silly). $A_n$, $n \geq 5$, doesn't have a subgroup of index 2, since such a subgroup would be normal and $A_n$ is simple (no non-trivial normal subgroups) for $n \geq 5$. 
This just leaves $n=4$, the case you're actually interested in. In this case $A_4$ would have to have a subgroup of order 6. It doesn't. Why not? Brute force check. :(
So in the end, the only subgroup of order $n!/2$ in $S_n$ is $A_n$ (for any $n$). 
A: HINT: $A_4$ is the group of even permutations in $S_4$. For any $x\in S_4$, $x^{-1}A_4x$ is a conjugate subgroup of $A_4$ and is also order 12. Can this subgroup have odd permutations? Can there be any subgroup of $S_4$ of order 12 with odd permutations? The minimal generating set for $A_4$ is 2 transpositions (odd $\times$ odd = even) and an even 3-cycle. Could an order 12 subgroup with an odd 3-cycle contain the identity (even permutation)?
Change in approach based upon whacka's comments. Thanks!
